Tangent cone to a subset of $\mathbb{R}^3$

Well, I have the set

$X=\{(x,y,z) \in \mathbb{R}^3 | 3x^2+2x^3+y^2+z^2=1\}$

How can I calculate the tangent cone at the point $(-1,0,0)$ ? What are the standard ways to calculate the tangent cone to a subset of $\mathbb{R}^3$ at a given point? Thank you!

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Well, one way would be to notice that you can set $z=0$ and work in the $(x,y)$-plane. There the cone will be just two lines and returning back to three dimensions, you obtain the full cone by revolving those two lines around their axis (since $y^2 + z^2 = r^2$ is an equation of a circle in the $(y,z)$-plane). – Marek Jan 24 '13 at 13:45
But how can I calculate the tangent cone to the subset $\{(x,y) \in \mathbb{R}^3 | 3x^2+2x^3+y^2=1\}$ at the point $(-1,0)$? – Frankenstein Jan 24 '13 at 14:27

The surface $X$ has rotational symmetry around the $x$-axis, therefore it is sufficient to work in the $(x,y)$-plane. Let me illustrate one way that we can find tangent lines.
Point $(-1,0)$ is critical. Near it, the curve will resemble couple of tangent lines. Let us therefore use the ansatz $y = kt$, $x = -1 + t$. We obtain $$\array{3(-1 + t)^2 + 2(-1 + t)^3 + k^2 t^2 = \cr 3 -6t + 3t^2 -2 +6t -6t^2 + 2t^3 + k^2 t^2 = \cr 1 + (k^2 - 3) t^2 + o(t^2)}$$ and this is supposed to be equal to $1$ as $t \to 0$. Therefore $k = \pm \sqrt{3}$. As a remark, note that in the above calculation the linear term vanished because $(-1,0)$ is critical.